Metamath Proof Explorer


Theorem ofdivrec

Description: Function analogue of divrec , a division analogue of ofnegsub . (Contributed by Steve Rodriguez, 3-Nov-2015)

Ref Expression
Assertion ofdivrec A V F : A G : A 0 F × f A × 1 ÷ f G = F ÷ f G

Proof

Step Hyp Ref Expression
1 simp1 A V F : A G : A 0 A V
2 simp2 A V F : A G : A 0 F : A
3 2 ffnd A V F : A G : A 0 F Fn A
4 ax-1cn 1
5 fnconstg 1 A × 1 Fn A
6 4 5 mp1i A V F : A G : A 0 A × 1 Fn A
7 simp3 A V F : A G : A 0 G : A 0
8 7 ffnd A V F : A G : A 0 G Fn A
9 inidm A A = A
10 6 8 1 1 9 offn A V F : A G : A 0 A × 1 ÷ f G Fn A
11 3 8 1 1 9 offn A V F : A G : A 0 F ÷ f G Fn A
12 eqidd A V F : A G : A 0 x A F x = F x
13 1cnd A V F : A G : A 0 1
14 eqidd A V F : A G : A 0 x A G x = G x
15 1 13 8 14 ofc1 A V F : A G : A 0 x A A × 1 ÷ f G x = 1 G x
16 ffvelrn F : A x A F x
17 2 16 sylan A V F : A G : A 0 x A F x
18 ffvelrn G : A 0 x A G x 0
19 eldifsn G x 0 G x G x 0
20 18 19 sylib G : A 0 x A G x G x 0
21 7 20 sylan A V F : A G : A 0 x A G x G x 0
22 divrec F x G x G x 0 F x G x = F x 1 G x
23 22 eqcomd F x G x G x 0 F x 1 G x = F x G x
24 23 3expb F x G x G x 0 F x 1 G x = F x G x
25 17 21 24 syl2anc A V F : A G : A 0 x A F x 1 G x = F x G x
26 3 8 1 1 9 12 14 ofval A V F : A G : A 0 x A F ÷ f G x = F x G x
27 25 26 eqtr4d A V F : A G : A 0 x A F x 1 G x = F ÷ f G x
28 1 3 10 11 12 15 27 offveq A V F : A G : A 0 F × f A × 1 ÷ f G = F ÷ f G